77
Acta Math. Univ. Comenianae
Vol. LXVIII, 1(1999), pp. 77–84
ON FINITE PRINCIPAL IDEAL RINGS
J. CAZARAN and A. V. KELAREV
Abstract. We find new conditions sufficient for a tensor product R ⊗ S and a
quotient ring Q/I to be a finite commutative principal ideal ring, where Q is a
polynomial ring and I is an ideal of Q generated by univariate polynomials.
1. Main Results
Finite commutative rings are interesting objects of ring theory and have many
applications in combinatorics. For these applications it is often important to know
when a ring is a principal ideal ring. Let us give only one example. Many classical
error-correcting codes are ideals in finite commutative rings. The existence of single
generators in ideals is important for computer storage as well as for encoding and
decoding algorithms (see [9]).
If we want to use certain ring constructions in combinatorial applications of
finite rings, then a natural question arises of when a ring construction is a principal
ideal ring. This question has been considered in the literature for several ring
constructions. For example, a complete description of commutative semigroup
rings which are PIR’s was obtained in [5]. All graded commutative principal ideal
rings were described in [4].
This paper is devoted to two ring constructions which are important, general
and lead to interesting results.
All rings considered are commutative and have identity elements. We write ⊗
for ⊗Z .
For any ring R and prime p, the p-component of R is defined by
Rp = {r ∈ R | pk r = 0 for some positive integer k}.
Let R be an arbitrary ring, p a prime, and let f ∈ R[x]. Denote by f the image
of f in R[x]/pR[x]. We say that f is squarefree (irreducible) modulo p if f
is squarefree (respectively, irreducible). A Galois ring GR(pm , r) is a ring of the
form (Z/pm Z)[x]/(f (x)), where p is a prime, m an integer, and f (x) ∈ Z/pm Z[x]
is a monic polynomial of degree r which is irreducible modulo p.
Received May 21, 1998.
1980 Mathematics Subject Classification (1991 Revision). Primary 13F10, 13F20.
The authors were supported by two grants of Australian Research Council.
78
J. CAZARAN and A. V. KELAREV
Theorem 1. A tensor product R ⊗ S of two finite commutative PIRs is a PIR
if and only if, for each prime p, at least one of the rings Rp and Sp is a direct
product of Galois rings.
Let R be a finite ring, Q = R[x1 , . . . , xn ] a polynomial ring. Our second main
theorem describes all rings of the form
R[x1 , . . . , xn ]/(f1 (x1 ), . . . , fn (xn ))
which are finite principal ideal rings. This gives a generalization of the main
result of [7]. Theorem 1 is used in the proof of Theorem 2. Ideals of the form
(f1 (x1 ), . . . , fn (xn )) are called elementary ideals (see [8, Definition 1.14]). A
few definitions are needed before we can state these results.
If F is a field, and f = g1m1 · · · gkmk , where f ∈ F [x] and g1 , . . . , gk are irreducible
polynomials over F , then by SP (f ) we denote the squarefree part g1 · · · gk of f .
We assume that SP (0) = 0.
Let R = GR(pm , r) = (Z/pm Z)[y]/(g(y)) 6= 0 be a Galois ring, which is not a
field. Then m > 1, because (Z/pZ)[y]/(g(y)) is a field, given that g(y) is irreducible
modulo p. We say that a polynomial f (x) ∈ R[x] is basic if all nonzero coefficients
of f (x) belong to the subset
B = {ay b | where 0 < a < p and 0 ≤ b < r}
of the Galois ring R, where r is the degree of g(y). Clearly, for every f ∈ R[x],
there exist unique basic polynomials
f 0 , f 00 ∈ B[x] ⊆ R[x] such that f − f 0 − pf 00 ∈ p2 R[x].
For any f ∈ R[x], there exists a unique basic polynomial SP (f ) ∈ R[x] such that
SP (f ) = SP (f ). Therefore there exists a unique basic polynomial UP (f ) ∈ R[x]
such that f = SP (f )UP (f ) or, equivalently, f 0 − SP (f ) UP (f ) ∈ pR[x]. Since f 0
is basic, (f 0 )00 = 0 for any f , and so (f 0 − SP (f ) UP (f ))00 = −(SP (f ) UP (f ))00 .
We introduce the following notation
fb = f 00 + (f 0 − SP (f ) UP (f ))00 = f 00 − (SP (f ) UP (f ))00 .
If the ideals of a ring form a chain, then it is called a chain ring (see
[6, p. 184]). By Lemma , every finite local principal ideal ring and every field
is a chain ring. A finite direct product is a PIR if and only if all its components
are PIRs (see [12, Theorem 33]). Since every finite PIR is a direct product of
chain rings (see [10, §6]), the general problem of describing all polynomial rings
Q = R[x1 , . . . , xn ]/(f1 (x1 ), . . . , fn (xn ))
which are finite PIRs reduces to the case where R is a chain ring. It follows from
[10, Theorem 13.2(c)], that Q is finite if and only if all the fi (xi ) are regular and
then we can assume that all the fi (xi ) are monic by [10, Theorem 13.6]. The
following theorem gives new conditions sufficient for Q to be a PIR.
ON FINITE PRINCIPAL IDEAL RINGS
79
Theorem 2. Let R be a finite commutative chain ring, and let f1 , . . . , fn be
univariate monic polynomials over R. Then
Q = R[x1 , . . . , xn ]/(f1 (x1 ), . . . , fn (xn ))
is a principal ideal ring and all rings R[xi ]/(fi (xi )) are PIRs, if one of the following
conditions is satisfied:
(i) R is a field and the number of polynomials fi which are not squarefree
does not exceed one;
(ii) R is a Galois ring of characteristic pm , for a prime p, the number of
polynomials f1 , . . . , fn which are not squarefree modulo p does not exceed
one, and if f = fi is not squarefree modulo p, then fb is coprime with
UP (f );
(iii) R is a chain ring, which is not a Galois ring, R has characteristic pm , for
a prime p, n = 1 and f1 is squarefree modulo p.
2. Proofs
The radical of a finite ring R is the largest nilpotent ideal N (R).
Lemma 3. A finite ring is a PIR if and only if its radical is a principal ideal.
Proof. The ‘only if’ part is trivial. If R is finite, then it is an Artinian ring.
Therefore it is a direct product of local rings ([1, Proposition 8.7]). If the radical of a local Artinian ring is a principal ideal, then all ideals are principal by
[1, Proposition 8.8].
Lemma 4. Let F be a finite field, P = F [x1 , . . . , xn ], and let I be the ideal
generated by f1 (x1 ), . . . , fn (xn ). Then the radical of P/I is equal to the ideal
generated by the squarefree parts of all polynomials f1 , . . . , fn .
Proof. Since every finite field is perfect, and any set of univariate polynomials
in pairwise distinct variables forms a Gröbner basis of the ideal it generates, this
lemma is a special case of more general results of [2, §8.2].
The ring GR(pn , r) is well defined independently of the monic polynomial of
degree r (see [10, §16]). Notice that GR(pm , 1) ∼
= Z/pm Z and GR(p, r) ∼
= GF (pr ),
r
n
the finite field of order p . For any f, g ∈ GR(p , r)[x], it is clear that f = g if
and only if f 0 = g 0 . The following lemma shows that a tensor product of Galois
rings is a PIR.
Lemma 5. ([10, Theorem 16.8]) Let p be a prime, k1 , k2 , r1 , r2 positive integers, and let k = min{k1 , k2 }, d = gcd(r1 , r2 ), m = lcm (r1 , r2 ). Then
GR(pk1 , r1 ) ⊗ GR(pk2 , r2 ) ∼
=
d
Y
1
GR(pk , m).
80
J. CAZARAN and A. V. KELAREV
In particular,
GF (p ) ⊗ GF (p ) ∼
=
r1
r2
d
Y
GF (pm ).
1
Lemma 6. ([10, Theorem 17.5]) Let R be a finite commutative ring which is
not a field. Then the following conditions are equivalent:
(i) R is a chain ring;
(ii) R is a local principal ideal ring;
(iii) there exist a prime p and integers m, r, n, s, t such that
R∼
= GR(pm , r)[x]/(g(x), pm−1 xt ),
where n is the index of nilpotency of the radical of R, t = n−(m−1)s > 0,
g(x) = xs + ph(x), deg(h) < s, and the constant term of h(x) is a unit in
GR(pm , r).
∼ GF (pr ).
Also, the characteristic of R is pm and its residue field is R/N (R) =
The polynomial g(x) which occurs in Lemma 6 is called an Eisenstein polynomial.
Lemma 7. Let R = GR(pm , r)[x]/(g(x), pm−1 xt ) be a chain ring, and let
s ≥ 2. Then the radical of R is generated by x.
Proof. Clearly, p is a nilpotent element of R. Therefore (x) is a nilpotent ideal,
because g(x) = xs + ph(x). Hence (x) ⊆ N (R). Given that g(x) = xs + ph(x) and
the constant term of h(x) is a unit in GR(pm , r), it follows that p ∈ (x). Since
R/(x) ∼
= GF (pr ) is a semisimple ring, we get (x) = N (R).
Lemma 8. ([10, Exercise 16.9]) A chain ring of characteristic pm is a Galois
ring if and only if its radical is generated by p. A PIR of characteristic pm is a
direct product of Galois rings if and only if its radical is generated by p.
Lemma 9. If R is a Galois ring, and S is a chain ring, then R ⊗ S is a PIR.
Proof. Let char (R) = pm , char (S) = q n , for primes p, q and positive integers
m, n. If p 6= q, then R ⊗ S = 0 is a PIR.
Suppose that p = q. Let g be the generator of the radical of S. Denote by (g) the
ideal generated by g in R⊗S. Clearly, (g) is nilpotent, and so (g) ⊆ N (R⊗S). It is
noted in the proof of Lemma 7 that p ∈ gS, and so p ∈ (g). Since S/gS ∼
= GF (pu )
and R/pR ∼
= GF (pv ), for some u, v, we get (R ⊗ S)/(g) ∼
= GF (pu ) ⊗ GF (pv ) ∼
=
Qd
w
GF
(p
)
where
w
=
lcm
{u,
v}
and
d
=
gcd{u,
v},
by
Lemma
5.
Therefore
1
(g) = N (R ⊗ S). By Lemma 3, R ⊗ S is a PIR.
ON FINITE PRINCIPAL IDEAL RINGS
81
Lemma 10. Let R and S be chain rings which are not Galois rings, and let
char (R) = pm , char (S) = pn , for a prime p and positive integers m, n. Then
R ⊗ S is not a PIR.
Proof. Suppose to the contrary that P = R ⊗ S is a PIR. Then P/pP is a
PIR, too. By Lemma 6 R ∼
= GR(pu , q)[x]/(xs + ph(x), pu−1 xt ). Since GR(pu , q)/
u
q
pGR(p , q) ∼
= GF (p ), we get R/pR ∼
= GF (pq )[x]/(xs ). If s = 1, then R =
GR(pu , q) is a Galois ring. Therefore s ≥ 2. Similarly, S/pS ∼
= GF (pr )[x]/(xt ),
q
2
for some t ≥ 2. It follows that H = GF (p )[x]/(x ) ⊗ GF (pr )[y]/(y 2 ) is a
homomorphic image of P/pP , and so H is a PIR. Further, H = (GF (pq ) ⊗
GF (pr ))[x, y]/(x2 , y 2 ). By Lemma 5 GF (pq ) ⊗ GF (pr ) is a direct product of finite
fields. Denote by F one of these fields. Then F [x, y]/(x2 , y 2 ) is a homomorphic
image of H, and so it is a PIR. However, if we set I = (x, y), then I is a maximal
ideal, and I 2 ⊂ (x2 , xy) ⊂ I. This is impossible by [6, Proposition 38.4(b)]. This
contradiction completes the proof.
Proof of Theorem 1. The ‘if’ part. Take any prime p. Suppose that Rp is a
direct product of Galois rings, and Sp is a PIR. Hence Sp is a direct product of
chain rings. Since tensor product distributes over direct products, Lemma 9 shows
that Rp ⊗ Sp is a PIR. Hence R ⊗ S is a PIR, because it is a direct product of a
finite number of rings Rp ⊗ Sp , for some p.
The ‘only if’ part. Given that R and S are PIRs, obviously Rp and Sp are
PIRs, for every p. Consider the decompositions of Rp and Sp into direct products
of chain rings. If both of these decompositions contain chain rings which are not
Galois rings, then we get a contradiction to Lemma 10. Thus at least one of the
rings Rp and Sp must be a product of Galois rings.
Lemma 11. Let R be a Galois ring of characteristic pm , f (x) a monic polynomial over R, and let Q = R[x]/(f (x)). Then Q is a direct product of Galois rings
if and only if f (x) is squarefree modulo p.
Proof. Lemma 4 shows that f (x) is squarefree modulo p if and only if Q/pQ is
semisimple, i.e., N (Q) = pQ. By Lemma 8 this is equivalent to Q being a direct
product of Galois rings.
Lemma 12. Let R = GR(pm , r) be a Galois ring, where m > 1, let f (x)
be a monic polynomial over R which is not squarefree modulo p, and let Q =
R[x]/(f (x)). Then Q is a PIR if UP (f ) is coprime with fb.
Proof. Given that f is not squarefree, we get UP (f ) 6= 0 and SP (f ) 6= 0.
Suppose that fb is coprime with UP (f ). Denote by h a basic polynomial in R[x]
b
such that h is the product of all irreducible divisors of f which do not divide f.
Let g = SP (f ) + ph ∈ R[x]. We claim that the radical N (Q) is equal to the ideal
I generated in Q by g.
82
J. CAZARAN and A. V. KELAREV
It follows from Lemma 4 that N (Q) = (SP (f ), p). Hence g ∈ N (Q), so I ⊆
N (Q). Therefore it remains to show that p, SP (f ) ∈ I.
First, we prove that pm−1 ∈ I. It suffices to show that pm−1 ∈ (g, f ) in
R[x], because I ⊆ Q = R[x]/(f ). The choice of h implies that fb − h UP (f )
is not divisible by any irreducible factor of f which does not divide fb. If an
b then it does not divide h, and so it does not
irreducible factor of f divides f,
b Thus fb − h UP (f ) and SP (f )
divide h UP (f ), because UP (f ) is coprime with f.
are coprime. Hence there exist basic polynomials v, w ∈ R[x] such that 1 =
v(fb − h UP (f )) + w SP (f ). There exists a unique basic polynomial f ∗ ∈ R[x]
satisfying f ∗ = fb. Since pm is the characteristic of R, pm u = 0 for all u ∈ R[x].
Therefore A = B is equivalent to pm−1 A = pm−1 B for all A, B ∈ R[x]. We can
lift the equation 1 = v(fb − h UP (f )) + w SP (f ) from R[x]/pR[x] ∼
= GF (pr )[x] to
m−1
R[x] and multiply by p
to get the following.
pm−1 = pm−1 [v(f ∗ − h UP (f )) + w SP (f )]
= pm−1 [v{f 00 + (f 0 − UP (f ) SP (f ))00 − h UP (f )} + w SP (f )]
= pm−2 [v{pf 00 + (f 0 − UP (f ) SP (f )) − ph UP (f )} + pw SP (f )]
= pm−2 [v(f 0 + pf 00 ) − v UP (f )(SP (f ) + ph) + pw SP (f )]
= pm−2 [vf − (v UP (f ) − pw)g] ∈ R[x].
We have used the fact that f 0 − UP (f ) SP (f ) = p[(f 0 − UP (f ) SP (f ))00 ] + p2 u
for some u ∈ R[x], because (f 0 − UP (f ) SP (f ))0 = 0. Thus pm−1 ∈ (g, f ) ⊂ R[x],
and so pm−1 ∈ I.
Since pm−1 belongs to both I and N (Q), we can factor out the ideal generated by pm−1 in Q and consider the ideal I/pm−1 I in Q/pm−1 Q. Also clearly
R/pm−1 R ∼
= GR(pm−1 , r). We identify f, g ∈ R[x] with their images in
(R/pm−1 R)[x]. We can now lift the equation 1 = v(fb − h UP (f )) + w SP (f )
from (R/pR)[x] to (R/pm−1 R)[x] and multiply by pm−2 and repeat the argument from the preceding paragraph taking into account that pm−1 u = 0 for all
u ∈ (R/pm−1 R)[x]. Then we deduce pm−2 ∈ (g, f ) ⊂ (R/pm−1 R)[x]. Identifying
pm−2 ∈ R[x] with its image pm−2 ∈ (R/pm−1 R)[x], we get pm−2 ∈ I/pm−1 I.
Given that pm−1 ∈ I, it follows that pm−2 ∈ I.
Repeating this reduction m − 3 times we get p ∈ I.
Next we prove that SP (f ) ∈ I. Since g, p ∈ I, then SP (f ) = g − ph ∈ I. Thus
I = N (Q), because N (Q) = (p, SP (f )). This means that N (Q) is a principal
ideal, and so Q is a PIR.
Lemma 13. Let R be a chain ring which is not a Galois ring, let f (x) be a
monic polynomial over R, and let Q = R[x]/(f (x)). Then Q is a PIR if and only
if f is squarefree modulo p.
Proof. By Lemma 6 R ∼
= GR(pm , r)[y]/(y s + ph(y), pm−1 y t ). Since R is not a
Galois ring, evidently s ≥ 2. Lemma 7 implies that p ∈ yR.
ON FINITE PRINCIPAL IDEAL RINGS
83
The ‘if’ part. Suppose that f is squarefree modulo p. Then Q/yQ ∼
= GF (pr )[x]/
(f ) is semisimple by Lemma 4. Thus N (Q) is a principal ideal. Lemma 3 tells us
that Q is a PIR.
The ‘only if’ part. Suppose that Q is a PIR then the ring Q/pQ ∼
= GF (pr )[x, y]/
s
(y , f (x)) is a PIR. This ring is isomorphic to the tensor product of GF (pr )[y]/(y s )
and GF (pr )[x]/(f (x)). Both of these rings are PIRs. Lemma 11 and Lemma 8 both
imply that GF (pr )[y]/(y s ) is not a direct product of Galois rings. By Lemma 8
GF (pr )[x]/(f (x)) must be a direct product of Galois rings. Lemma 11 completes
the proof.
Proof of Theorem 2. The ring Q is isomorphic to the tensor product of the rings
R[xi ]/(fi (xi )), for i = 1, . . . , n.
(i): Suppose that R is a field of characteristic p. Then all the R[xi ]/(fi (xi )) are
PIRs. Theorem 1 tells us that Q is a PIR if and only if at least n − 1 of the rings
R[xi ]/(fi (xi )) are direct products of Galois rings. By Lemma 11 this is equivalent
to the fact that at most one of the polynomials fi (xi ) is not squarefree.
(ii): Suppose that R is a Galois ring. By Lemma 12 all R[xi ]/(fi (xi )) are PIRs
if, for each polynomial fi (xi ) which is not squarefree modulo p, UP (fi ) is coprime
with fbi . Further, suppose that this condition is satisfied. As in case (i), we see
that Q is a PIR if at most one of the polynomials fi (xi ) is not squarefree modulo p.
(iii): Suppose that R is a chain ring which is not a Galois ring. Since the
class of finite direct products of Galois rings is closed for homomorphic images by
Lemma 8, we see that each R[xi ]/(fi (xi )) is not a direct product of Galois rings.
Theorem 1 shows that n = 1. By Lemma 13 Q is a PIR if and only if f1 (x1 ) is
squarefree modulo p.
For finite rings, our Theorem 2 immediately gives the following Theorem 1
of [7].
Corollary 14. ([7]) Let F be a field of characteristic p > 0, a1 , . . . , an nonnegative integers, b1 , . . . , bn positive integers, and let
R = F [x1 , . . . , xn ]/(xa1 1 (1 − xb11 ), . . . , xann (1 − xbnn )).
then R is a principal ideal ring if and only if one of the following conditions is
satisfied:
(1) a1 , . . . , an ≤ 1 and p divides at most one number among b1 , . . . , bn ;
(2) exactly one of a1 , . . . , an , say a1 , is greater than 1 and p does not divide
each of b2 , . . . , bn .
Proof. Consider the polynomial f = xa (1 − xb ). By [2, Lemma 2.85], a polynomial is squarefree if and only if it is coprime with its derivative. Since char F =
p > 0, then f is squarefree if and only if a = 1 and p does not divide b. Thus
Theorem 2 completes the proof.
84
J. CAZARAN and A. V. KELAREV
References
1. Atiyah M. and McDonald I., Introduction to Commutative Algebra, Addison-Wesley Pub.
Co., 1969.
2. Becker T. and Weispfenning V., Gröbner Bases. A Computational Approach to Commutative
Algebra, Springer-Verlag, 1993.
3. Cazaran J. and Kelarev A. V., Generators and weights of polynomial codes, Archiv Math.
(Basel) 69 (1997), 479–486.
4. Decruyenaere F. and Jespers E., Graded Commutative Principal Ideal Rings, Bull. Belg.
Math. Soc. Ser. B 43 (1991), 143–150.
5. Decruyenaere F., Jespers E. and Wauters P., On Commutative Principal Ideal Semigroup
Rings, Semigroup Forum 43 (1991), 367–377.
6. Gilmer R., Multiplicative Ideal Theory, Marcel Dekker Inc., New York, 1972.
7. Glastad B. and Hopkins G., Commutative semigroup rings which are principal ideal rings,
Comment. Math. Univ. Carolinae 21 (1980), 371–377.
8. Kurakin V. L., Kuzmin A. S., Mikhalev A. V. and Nechaev A. A., Linear recurring sequences
over rings and modules, Journal of Mathematical Sciences 76(6) (1995), 2793–2915.
9. Landrock P. and Manz O., Classical codes as ideals in group algebras, Des. Codes Cryptogr.
2(3) (1992), 273–285.
10. McDonald B. R., Finite Rings with Identity, Marcel Dekker, New York, 1974.
11. Nagata M., Local rings, John Wiley & Sons, New York, 1962.
12. Zariski O. and Samuel P., Commutative Algebra, Van Nostrand, Princeton, New Jersey, 1958.
J. Cazaran, Department of Mathematics, University of Tasmania, G.P.O. Box 252-37, Hobart,
Tasmania 7001, Australia; e-mail: cazaran@hilbert.maths.utas.edu.au
A. V. Kelarev, Department of Mathematics, University of Tasmania, G.P.O. Box 252-37, Hobart,
Tasmania 7001, Australia; e-mail: kelarev@hilbert.maths.utas.edu.au